Verified answer

From your previous question,

∫ [ 2x / (2x + 1 - x^2) ]dx =

∫ [- 2x / (x - 1 - √2)(x - 1 + √2) ]dx =

∫ [- 2(1 - √2) / (x - 1 - √2) - 2(1 + √2) / (x - 1 + √2) ]dx =

- 2∫ [(1 - √2) / (x - 1 - √2) + (1 + √2) / (x - 1 + √2) ]dx =

- 2 [ (1 - √2) Ln (x - 1 - √2) + (1 + √2) Ln (x - 1 + √2) ] + C =

- 2 [ (Ln (x - 1 - √2) - √2Ln (x - 1 - √2)) + (Ln (x - 1 + √2) + √2Ln (x - 1 + √2)) ] + C =

- 2 [ Ln (x - 1 - √2) + Ln (x - 1 + √2) + √2Ln (x - 1 + √2) - √2Ln (x - 1 - √2) ] + C =

- 2 [ Ln (x^2 - 2x - 1) + √2Ln [(x - 1 + √2) / (x - 1 - √2) ] ] + C

This can be "simplified" still further . . .

(1/2)((-2 + sqr(2))Log(-1 + sqr(2) + x) - (2+sqr(2))Log(1+sqr(2) -x))

Looks complicated I know but if you differentiate it it will work, promise :)